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Suppose that $\mathbf A=\mathbf A(t)$ is a matrix whose entries are parametrized by a variable $t$. The Jacobi's formular states that $$ \frac d{dt}\left( \det \mathbf A\right)= \text{Tr}\left( \text{adj} (\mathbf A ) \frac{d\mathbf A}{dt} \right)\ , $$ where $\text{adj}(\mathbf A)$ is the adjugate (or adjoint) of $\mathbf A$. There is a proof of that in this wikipedia article. Another closely related identity, which appears to go by the same name, is $$ \det\left( e^{\mathbf A} \right) = e^{\text{Tr}(\mathbf A)}\ . $$ I want to understand the reason why these formulas are true, especially the second one. In particularly, what is the interpretations of $\det\left( e^{\mathbf A} \right)$ and $\text{Tr}(\mathbf A)$, when $\mathbf A$ is considered as a linear operator?

These are some related links that I have found about the question:

Derivative of a determinant of a matrix field

Proof for the derivative of the determinant of a matrix

Exponential of a matrix and related derivative

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  • $\begingroup$ $Tr (A)$ is the trace of a matrix, while $e^A = I +A + A^2/2! + \cdots + A^n /n! + \cdots$ $\endgroup$ – user99914 May 12 '16 at 2:26
  • $\begingroup$ @JohnMa I know what they mean... What I want to know is their intepretation, like $\det(\mathbf A)$ represents the $n-$dimensional volume of the image of a unit $n-$cube under $\mathbf A$. $\endgroup$ – BigbearZzz May 12 '16 at 2:34
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    $\begingroup$ Maybe not what you're looking for, but the second equation can be proved pretty easily for diagonalizable $A$ $\det( e^A ) = \prod_i e^{\lambda_i} = e^{\sum_i \lambda_i} = e^{\mathrm{Tr}(A)}$ $\endgroup$ – D.A.N. May 12 '16 at 2:51
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When interpreting the exponential for a linear map $f:V\to V$ on a finite dimensional real vector space $V$, it seems better to me not to look at $e^f$ but at the family $e^{tf}$ for $t\in\mathbb R$. This is the solution operator for the linear first order ODE $x'=f(x)$, i.e. the unique solution with initial condition $x(0)=x_0$ is given by $x(t)=e^{tf}(x_0)$. Hence $e^f$ can be interpreted as computing the dependence of the solution at time $t=1$ on the initial condition. At this stage you can use the interpretation of the determinant via volumes you suggested.

Now the properties of the exponential readily imply that $t\mapsto \det(e^{tf})=:\phi(t)$ satisfies $\phi(t+s)=\phi(t)\phi(s)$ and $\phi(0)=1$, which easily implies that $\phi'(t)=\lambda\phi(t)$, where $\lambda=\phi'(0)$. The unique solution of this ODE with initial condition $\phi(0)=1$ is $\phi(t)=e^{\lambda t}$. Hence we conclude that $\det(e^{tf})=e^{t\lambda(f)}$ for some function $\lambda:L(V,V)\to\mathbb R$.

It is also not difficult to see that $\lambda$ has to be invariant under conjugation: Observe that $c(t)=e^{tf}$ is uniquely determined by $c(0)=id$ and $c'(t)=f\circ c(t)$. Now if $g:V\to V$ is a linear isomorphism, it is easy to conclude that $e^{t(g\circ f\circ g^{-1})}=g\circ e^{tf}\circ g^{-1}$. Taking determinants, you conclude that $\lambda(g\circ f\circ g^{-1})=\lambda(f)$. With a bit more effort, you can show that $\lambda$ is linear, which implies that it is a multiple of the trace. Checking the case $f=id$ then sorts out that $\lambda(f)=tr(f)$. Unfortunately, I am not aware of an "intuitive" interpretation of the trace in this setting, apart that it can be viewed as computing the unique $GL(V)$-invariant projection of $f$ onto multiples of the identity.

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